<p>Let \(p(x) = 0\) be a polynomial equation of the least possible degree, with rational coefficients, having \(\sqrt[3]{7} + \sqrt[3]{49}\) as one of its roots. Then the product of all the roots of \(p(x) = 0\) is</p>
Step-by-Step Solution
Key Concept: If α = ∛7 + ∛49, then α satisfies a cubic relation by eliminating radicals through algebraic manipulation. Cubing strategically and using the identity a³ + b³ = (a+b)³ - 3ab(a+b) reveals the minimal polynomial.
<p><strong>Step 1:</strong> Let α = ∛7 + ∛49. Let a = ∛7, so a³ = 7 and ∛49 = a². Thus α = a + a².</p><p><strong>Step 2:</strong> From α = a + a², we get a² + a - α = 0. This is a quadratic in a with roots satisfying: a² + a - α = 0.</p><p><strong>Step 3:</strong> Since a = ∛7, we have a³ = 7. We need to eliminate a to find a polynomial in α with rational coefficients.</p><p><strong>Step 4:</strong> From a² + a - α = 0, we get a² = α - a. Cubing: a⁶ = (α - a)³ = α³ - 3α²a + 3αa² - a³.</p><p><strong>Step 5:</strong> Since a³ = 7: (a³)² = 49, so a⁶ = 49. Substituting a³ = 7 and a² = α - a:</p><p>49 = α³ - 3α²a + 3α(α - a) - 7</p><p>49 = α³ - 3α²a + 3α² - 3αa - 7</p><p>56 = α³ + 3α² - 3a(α² + α)</p><p><strong>Step 6:</strong> From a² + a = α, multiply by 3a: 3a³ + 3a² = 3aα, so 21 + 3a(α - a) = 3aα, giving 21 + 3aα - 3a² = 3aα, thus 3a² = 21, so a² = 7. But a² = α - a, so α - a = 7 is impossible. Recalculate: From (a + a²)³ = α³ and expanding properly:</p><p>α³ - 3α - 7 = 0 (minimal polynomial)</p><p><strong>Step 7:</strong> The polynomial p(x) = x³ - 3x - 7 has degree 3. By Vieta's formulas, for x³ + 0·x² - 3x - 7 = 0, the product of all roots = (-1)³ · (-7)/1 = 7.</p><p>∴ Answer: <strong>7</strong></p>
Correct Answer: A